Question: Evaluate $~~\int x^2\ln x\ dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{x^3\,\ln x-x^3}3+C$ (Choice B) B $\dfrac{x^3\,\ln x}3-\dfrac19x^3+C$ (Choice C) C $\dfrac16x^3+C$ (Choice D) D $\dfrac{x^3\,\ln x+x^3}3+C$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv=x^2\ dx\,$. Then $~du = \dfrac1xdx~$ and $~v = \int x^2dx = \dfrac13x^3\,$. Integration by parts gives $ \int x^2\ln xdx = \ln x\cdot\dfrac13x^3-\int\dfrac13x^3\cdot\dfrac1xdx$ $ ~~\,=\dfrac{x^3\,\ln x }3-\int\dfrac{x^2}3dx$ $ ~~\,=\dfrac{x^3\,\ln x}3-\dfrac19x^3+C\,$.